Chapter 17: Time-Independent Perturbation Theory — The Non-Degenerate Case

"Give me an unperturbed problem I can solve exactly, and I will give you an approximate answer to any problem in its neighborhood." — The implicit promise of perturbation theory

We have arrived at a turning point in this textbook. For sixteen chapters, we have been building an increasingly powerful mathematical apparatus — wavefunctions, operators, Dirac notation, eigenvalue problems, angular momentum, spin, identical particles — and applying it to problems that can be solved exactly. The infinite square well. The quantum harmonic oscillator. The hydrogen atom. These are beautiful systems, and the fact that quantum mechanics yields closed-form solutions for them is one of the triumphs of twentieth-century physics.

But here is the uncomfortable truth: almost nothing in nature corresponds to an exactly solvable problem. Real atoms sit in external electric and magnetic fields. Real molecules have bonds that are not perfectly parabolic. Real solids have impurities, defects, and interactions between electrons that defy closed-form analysis. If quantum mechanics could only solve the handful of textbook problems we have encountered so far, it would be a mathematical curiosity rather than the most successful physical theory ever constructed.

This chapter introduces the tool that bridges the gap between the exactly solvable and the physically real: perturbation theory. The idea is simple, even if the execution requires care. We start with a problem we can solve — the unperturbed system — and then systematically account for the effects of a small additional term in the Hamiltonian. The result is a series expansion for energies and states that can be carried to whatever order of accuracy we require.

💡 Key Insight — Perturbation theory is not a sign of weakness. It is the primary computational tool of professional physicists. The Standard Model of particle physics, the electronic structure of materials, the properties of lasers — all of these rely on perturbative calculations. When someone says they "solved" a quantum problem, they almost always mean they computed the first few orders of a perturbation expansion.

Learning paths: - 🏃 Streamlined path: Sections 17.1–17.5 are essential for every subsequent chapter. Section 17.6 (anharmonic oscillator) is the must-do application. Skim Section 17.7 (Stark effect) on first reading if pressed for time, but return to it before Chapter 18. - 🔬 Deep dive path: Work through everything sequentially. Section 17.8 on convergence is where the real subtlety lives and connects to deep questions about the structure of quantum mechanics.

17.1 Why Approximation Methods Matter

The Landscape of Solvable Problems

Let us take stock of what we can solve exactly. In all of quantum mechanics, the list of exactly solvable potentials is remarkably short:

That is essentially it. Six problems (with variations). Everything else in quantum mechanics — the helium atom, the hydrogen atom in an electric field, the anharmonic oscillator, molecular bonding, band structure in solids — requires approximation.

🔗 Connection — In Chapter 4, we solved the quantum harmonic oscillator exactly. In this chapter, we will deliberately break it by adding an anharmonic term $\lambda x^4$, and then use perturbation theory to recover approximate answers. The QHO is our launching pad into the real world.

Three Pillars of Approximation

Quantum mechanics offers three major approximation frameworks, each with its own domain of applicability:

1. Perturbation theory (this chapter and Chapters 18, 21): Best when the problem is close to an exactly solvable one. We have a small parameter $\lambda$ that measures how far we are from the solvable case.

2. The variational method (Chapter 19): Best when we want an upper bound on the ground-state energy and do not need a systematic expansion. We guess a trial wavefunction and optimize.

3. The WKB approximation (Chapter 20): Best for semi-classical problems where the potential varies slowly compared to the wavelength.

Each method has strengths and limitations. Perturbation theory is the most versatile and the most widely used. It is also the most dangerous, because it can give misleading results when applied outside its domain of validity — a topic we address head-on in Section 17.8.

⚠️ Common Misconception — Students often think that "approximate" means "inaccurate." This is wrong. The perturbative calculation of the electron's magnetic moment ($g$-factor) agrees with experiment to twelve significant figures — making it the most precisely verified prediction in all of science. Approximation methods, used wisely, are extraordinarily precise.

The Art of Choosing the Unperturbed System

Before we develop the formalism, it is worth emphasizing a point that textbooks often gloss over: the decomposition $\hat{H} = \hat{H}_0 + \hat{H}'$ is not unique. There are infinitely many ways to split a Hamiltonian into "solvable" and "perturbation" parts, and the quality of the perturbative approximation depends sensitively on this choice.

Consider the helium atom. The Hamiltonian is:

$$\hat{H} = -\frac{\hbar^2}{2m_e}\nabla_1^2 - \frac{\hbar^2}{2m_e}\nabla_2^2 - \frac{2e^2}{4\pi\epsilon_0 r_1} - \frac{2e^2}{4\pi\epsilon_0 r_2} + \frac{e^2}{4\pi\epsilon_0 r_{12}}$$

The "obvious" choice is $\hat{H}_0 = $ two independent hydrogen-like atoms with $Z=2$, and $\hat{H}' = e^2/(4\pi\epsilon_0 r_{12})$ (electron-electron repulsion). This gives first-order results within 5% of experiment.

But a better choice might be $\hat{H}_0 = $ two independent electrons in an effective potential $V(r) = -Z_{\text{eff}}e^2/(4\pi\epsilon_0 r)$ with $Z_{\text{eff}} \approx 1.69$ (chosen to account for screening), and $\hat{H}' = $ the difference between the true Hamiltonian and $\hat{H}_0$. The perturbation is smaller, and the first-order results are more accurate.

The lesson: perturbation theory works best when $\hat{H}_0$ already captures the essential physics. The perturbation should be a refinement, not a fundamental ingredient.

🔵 Historical Note — The idea that one should choose $\hat{H}_0$ wisely was formalized by Brillouin and Wigner in the 1930s. Their approach (Brillouin-Wigner perturbation theory) differs subtly from the Rayleigh-Schrödinger approach we develop here, but the philosophical point is the same. In modern many-body physics, the choice of unperturbed Hamiltonian — often called the "reference state" — is elevated to an art form in methods like Hartree-Fock, density functional theory, and coupled-cluster theory.

17.2 The Perturbation Expansion: $\hat{H} = \hat{H}_0 + \lambda\hat{H}'$

Setting Up the Framework

We begin with a Hamiltonian that can be split into two pieces:

$$\hat{H} = \hat{H}_0 + \lambda\hat{H}' \tag{17.1}$$

Here: - $\hat{H}_0$ is the unperturbed Hamiltonian, whose eigenvalues and eigenstates we know exactly:

$$\hat{H}_0|n^{(0)}\rangle = E_n^{(0)}|n^{(0)}\rangle \tag{17.2}$$

• $\hat{H}'$ is the perturbation — the additional piece that makes the problem unsolvable in closed form.

• $\lambda$ is a dimensionless perturbation parameter satisfying $0 \le \lambda \le 1$. At $\lambda = 0$, we recover the unperturbed problem. At $\lambda = 1$, we have the full physical problem. The role of $\lambda$ is to serve as a bookkeeping device that tracks the order of each correction.

The unperturbed states $\{|n^{(0)}\rangle\}$ form a complete orthonormal set:

$$\langle m^{(0)}|n^{(0)}\rangle = \delta_{mn}, \qquad \sum_n |n^{(0)}\rangle\langle n^{(0)}| = \hat{1} \tag{17.3}$$

🔗 Connection — The completeness relation (17.3) was introduced in Chapter 8 and developed extensively in Chapter 9. If this feels unfamiliar, revisit Section 9.3 before proceeding. The entire perturbation expansion relies on expanding unknown states in the complete basis of known states.

The Non-Degeneracy Assumption

In this chapter, we restrict ourselves to the case where the unperturbed energy levels are non-degenerate: no two different states share the same energy. Formally:

$$E_n^{(0)} \neq E_m^{(0)} \quad \text{for } n \neq m \tag{17.4}$$

This is crucial. When degeneracy is present — as in the hydrogen atom, where all states with the same principal quantum number $n$ have the same energy — the formulas we derive here break down spectacularly (denominators go to zero). The degenerate case requires its own treatment, which we develop in Chapter 18.

🔴 Warning — If you encounter a perturbation problem and blithely apply the non-degenerate formulas without checking for degeneracy, you will get infinities. Always verify that the unperturbed spectrum is non-degenerate before using the results of this chapter.

The Power Series Ansatz

We now make the key assumption: the exact energies and eigenstates of the full Hamiltonian $\hat{H}$ can be expanded in powers of $\lambda$:

$$E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \lambda^3 E_n^{(3)} + \cdots \tag{17.5}$$

$$|n\rangle = |n^{(0)}\rangle + \lambda|n^{(1)}\rangle + \lambda^2|n^{(2)}\rangle + \cdots \tag{17.6}$$

The superscript in parentheses denotes the order of the correction: - $E_n^{(0)}$ is the zeroth-order (unperturbed) energy. - $E_n^{(1)}$ is the first-order energy correction — proportional to the strength of the perturbation. - $E_n^{(2)}$ is the second-order correction — proportional to the square of the perturbation strength. - $|n^{(1)}\rangle$ is the first-order correction to the state, and so on.

The idea is that if $\lambda$ is small, the series converges rapidly and we can truncate after a few terms. (Whether the series actually converges is a subtle question we tackle in Section 17.8.)

Deriving the Master Equations

Substituting the expansions (17.5) and (17.6) into the eigenvalue equation $\hat{H}|n\rangle = E_n|n\rangle$:

$$(\hat{H}_0 + \lambda\hat{H}')(|n^{(0)}\rangle + \lambda|n^{(1)}\rangle + \lambda^2|n^{(2)}\rangle + \cdots) = (E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \cdots)(|n^{(0)}\rangle + \lambda|n^{(1)}\rangle + \cdots)$$

Now we collect terms by powers of $\lambda$. Since this equation must hold for all values of $\lambda$, the coefficients of each power of $\lambda$ must separately be equal.

Zeroth order ($\lambda^0$):

$$\hat{H}_0|n^{(0)}\rangle = E_n^{(0)}|n^{(0)}\rangle \tag{17.7}$$

This is just the unperturbed eigenvalue equation — it is satisfied by construction.

First order ($\lambda^1$):

$$\hat{H}_0|n^{(1)}\rangle + \hat{H}'|n^{(0)}\rangle = E_n^{(0)}|n^{(1)}\rangle + E_n^{(1)}|n^{(0)}\rangle \tag{17.8}$$

Second order ($\lambda^2$):

$$\hat{H}_0|n^{(2)}\rangle + \hat{H}'|n^{(1)}\rangle = E_n^{(0)}|n^{(2)}\rangle + E_n^{(1)}|n^{(1)}\rangle + E_n^{(2)}|n^{(0)}\rangle \tag{17.9}$$

The pattern continues to all orders. Each order equation relates the correction at that order to quantities from lower orders.

Normalization Convention

We need a convention for normalizing the perturbed states. The most convenient choice is intermediate normalization:

$$\langle n^{(0)}|n\rangle = 1 \tag{17.10}$$

This means the perturbed state $|n\rangle$ is not normalized to unity, but its overlap with the unperturbed state is exactly 1. From expansion (17.6), this requires:

$$\langle n^{(0)}|n^{(k)}\rangle = 0 \quad \text{for all } k \ge 1 \tag{17.11}$$

In other words, the correction terms $|n^{(1)}\rangle, |n^{(2)}\rangle, \ldots$ are all orthogonal to the unperturbed state. This makes physical sense: the corrections represent the admixture of other unperturbed states into the original state due to the perturbation.

Higher Orders: The Pattern

The pattern should now be clear. At each order $k$, we get one equation that relates $|n^{(k)}\rangle$ and $E_n^{(k)}$ to quantities from all lower orders. The $k$-th order equation is:

$$\hat{H}_0|n^{(k)}\rangle + \hat{H}'|n^{(k-1)}\rangle = \sum_{j=0}^{k} E_n^{(j)}|n^{(k-j)}\rangle \tag{17.9b}$$

Acting from the left with $\langle n^{(0)}|$ extracts $E_n^{(k)}$, while acting with $\langle m^{(0)}|$ ($m \neq n$) extracts the coefficients of $|n^{(k)}\rangle$. This recursive structure means that, in principle, we can compute corrections to arbitrary order — though in practice the algebra becomes prohibitive beyond the first few orders without computer algebra systems.

🧪 Experiment — In quantum electrodynamics, the perturbation parameter is the fine-structure constant $\alpha \approx 1/137$. The anomalous magnetic moment of the electron has been computed to fifth order in $\alpha$ (that is, through $\alpha^5$, involving the evaluation of 12,672 Feynman diagrams). The agreement with experiment — $g/2 = 1.001\,159\,652\,180\,73(28)$ (theory) vs. $1.001\,159\,652\,180\,59(13)$ (experiment) — is one of the greatest achievements of theoretical physics. This is perturbation theory in action, pushed to extraordinary precision.

✅ Checkpoint — Before proceeding, make sure you can: - Write the perturbation expansion for $E_n$ and $|n\rangle$ to second order - Explain why the first-order equation (17.8) involves both $|n^{(1)}\rangle$ and $E_n^{(1)}$ - State the non-degeneracy condition and explain why it matters

17.3 First-Order Energy Correction: $E_n^{(1)} = \langle n^{(0)}|\hat{H}'|n^{(0)}\rangle$

The Derivation

We extract the first-order energy correction from equation (17.8) by acting on the left with $\langle n^{(0)}|$:

$$\langle n^{(0)}|\hat{H}_0|n^{(1)}\rangle + \langle n^{(0)}|\hat{H}'|n^{(0)}\rangle = E_n^{(0)}\langle n^{(0)}|n^{(1)}\rangle + E_n^{(1)}\langle n^{(0)}|n^{(0)}\rangle$$

Now we use three facts: 1. $\langle n^{(0)}|\hat{H}_0 = E_n^{(0)}\langle n^{(0)}|$ (because $\hat{H}_0$ is Hermitian and $|n^{(0)}\rangle$ is an eigenstate). 2. $\langle n^{(0)}|n^{(1)}\rangle = 0$ from our normalization convention (17.11). 3. $\langle n^{(0)}|n^{(0)}\rangle = 1$ (unperturbed states are normalized).

The first term on the left becomes $E_n^{(0)}\langle n^{(0)}|n^{(1)}\rangle = 0$, and the first term on the right also vanishes. We are left with:

$$\boxed{E_n^{(1)} = \langle n^{(0)}|\hat{H}'|n^{(0)}\rangle} \tag{17.12}$$

This is one of the most important results in all of quantum mechanics. Let us state it plainly:

The first-order energy correction is the expectation value of the perturbation in the unperturbed state.

💡 Key Insight — Equation (17.12) has a beautiful physical interpretation. To first approximation, the perturbation shifts the energy of state $|n^{(0)}\rangle$ by exactly the amount you would compute if the particle "stayed in" the unperturbed state and simply experienced the perturbation as an average field. The state has not changed yet — only the energy has shifted. The wavefunction corrections (which capture the actual redistribution of probability) enter at the same order, but the energy formula only needs the unperturbed state.

Properties of the First-Order Correction

Several important properties follow immediately:

Linearity. If the perturbation is a sum $\hat{H}' = \hat{V}_1 + \hat{V}_2$, then $E_n^{(1)} = \langle n^{(0)}|\hat{V}_1|n^{(0)}\rangle + \langle n^{(0)}|\hat{V}_2|n^{(0)}\rangle$. The first-order corrections from different perturbations simply add.

Variational bound. For the ground state ($n = 0$), the first-order corrected energy $E_0^{(0)} + E_0^{(1)}$ is not guaranteed to be an upper bound. This is a key difference from the variational method (Chapter 19), which always gives an upper bound.

Symmetry selection rules. If the unperturbed state has definite parity (even or odd) and the perturbation has odd parity, then $E_n^{(1)} = 0$. This is because the integrand $\langle n^{(0)}|\hat{H}'|n^{(0)}\rangle$ involves the product of an even function, an odd function, and an even function, which is odd overall and integrates to zero.

📊 By the Numbers — The first-order correction to the ground state of helium (treating the electron-electron repulsion as the perturbation) gives $E_0^{(1)} = \frac{5}{4}Z \cdot E_1^{(H)} = -\frac{5}{4}(2)(13.6\,\text{eV}) = -34\,\text{eV}$. The unperturbed energy (two non-interacting electrons in the Coulomb field of $Z=2$) is $E_0^{(0)} = 2 \times (-Z^2)(13.6\,\text{eV}) = -108.8\,\text{eV}$. The total first-order estimate is $-108.8 + 34.0 = -74.8\,\text{eV}$, compared to the experimental value of $-78.975\,\text{eV}$. This is already within 5% — not bad for a one-line calculation.

A Simple Warm-Up: Tilted Infinite Square Well

As a first application, consider a particle in an infinite square well ($0 < x < L$) with a small linear perturbation:

$$\hat{H}' = V_0 \frac{x}{L}$$

This tilts the flat bottom of the well. The unperturbed energies and states are:

$$E_n^{(0)} = \frac{n^2\pi^2\hbar^2}{2mL^2}, \qquad \langle x|n^{(0)}\rangle = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$

The first-order correction is:

$$E_n^{(1)} = \langle n^{(0)}|\hat{H}'|n^{(0)}\rangle = \frac{2V_0}{L^2}\int_0^L x\sin^2\left(\frac{n\pi x}{L}\right)dx = \frac{V_0}{2}$$

The integral evaluates to $L/2$ for all $n$ (this can be shown using the identity $\sin^2\theta = (1-\cos 2\theta)/2$). The physical interpretation is transparent: the average position of the particle is at the center of the well for any state, and the average perturbation energy is $V_0 \cdot (L/2)/L = V_0/2$.

✅ Checkpoint — Verify for yourself that $\int_0^L x \sin^2(n\pi x/L)\,dx = L^2/4$ for any positive integer $n$. (Hint: use integration by parts after applying the double-angle formula.)

Another Example: The Harmonic Oscillator with a Linear Perturbation

As a second warm-up example, consider the QHO with a small constant force:

$$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2 + F\hat{x}$$

where $F$ is a constant force. The perturbation is $\hat{H}' = F\hat{x}$. Using the ladder operator expression $\hat{x} = \sqrt{\hbar/(2m\omega)}(\hat{a}+\hat{a}^\dagger)$:

$$E_n^{(1)} = \langle n|\hat{H}'|n\rangle = F\sqrt{\frac{\hbar}{2m\omega}}\langle n|(\hat{a}+\hat{a}^\dagger)|n\rangle = 0$$

since $\hat{a}|n\rangle = \sqrt{n}|n-1\rangle$ and $\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$ are both orthogonal to $|n\rangle$.

This result is also obvious from symmetry: the QHO states are centered at $x = 0$ (for even $n$, the probability density is symmetric about the origin; for odd $n$ it is antisymmetric, but $|\psi|^2$ is still symmetric). Therefore $\langle x\rangle = 0$ for all states, and the average force does no work.

But we know this problem can be solved exactly — by completing the square:

$$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\left(\hat{x} + \frac{F}{m\omega^2}\right)^2 - \frac{F^2}{2m\omega^2}$$

This is just a harmonic oscillator shifted to a new equilibrium position $x_0 = -F/(m\omega^2)$, with all energies shifted down by $F^2/(2m\omega^2)$:

$$E_n = \left(n + \frac{1}{2}\right)\hbar\omega - \frac{F^2}{2m\omega^2}$$

The first-order correction vanishes (as we computed), and the entire effect is captured by the second-order correction $E_n^{(2)} = -F^2/(2m\omega^2)$. In fact, the second-order correction is exact — all higher-order corrections vanish identically. This is a special property of the linear perturbation on the harmonic oscillator: because $\hat{x}$ connects only neighboring states ($|n\rangle \to |n\pm 1\rangle$), and each state appears at most once in the perturbation chain, the expansion terminates at finite order.

⚠️ Common Misconception — The vanishing of the first-order correction does NOT mean the perturbation has no effect. It means the leading effect appears at second order. In the Stark effect (Section 17.7), we will see the same phenomenon: the first-order correction vanishes by parity, and the physical effect (polarizability) appears at second order.

17.4 First-Order Wavefunction Correction

Expanding in the Unperturbed Basis

To find $|n^{(1)}\rangle$, we return to the first-order equation (17.8) and exploit the completeness of $\{|m^{(0)}\rangle\}$. Since $|n^{(1)}\rangle$ is orthogonal to $|n^{(0)}\rangle$ (by our normalization convention), we can expand it in terms of all the other unperturbed states:

$$|n^{(1)}\rangle = \sum_{m \neq n} c_m^{(1)} |m^{(0)}\rangle \tag{17.13}$$

To find the coefficients $c_m^{(1)}$, we act on the first-order equation (17.8) from the left with $\langle m^{(0)}|$ for $m \neq n$:

$$\langle m^{(0)}|\hat{H}_0|n^{(1)}\rangle + \langle m^{(0)}|\hat{H}'|n^{(0)}\rangle = E_n^{(0)}\langle m^{(0)}|n^{(1)}\rangle + E_n^{(1)}\langle m^{(0)}|n^{(0)}\rangle$$

The last term vanishes because $\langle m^{(0)}|n^{(0)}\rangle = \delta_{mn} = 0$ for $m \neq n$. Using $\langle m^{(0)}|\hat{H}_0 = E_m^{(0)}\langle m^{(0)}|$ and $\langle m^{(0)}|n^{(1)}\rangle = c_m^{(1)}$:

$$E_m^{(0)} c_m^{(1)} + \langle m^{(0)}|\hat{H}'|n^{(0)}\rangle = E_n^{(0)} c_m^{(1)}$$

Solving for $c_m^{(1)}$:

$$c_m^{(1)} = \frac{\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle}{E_n^{(0)} - E_m^{(0)}} \tag{17.14}$$

Therefore, the first-order wavefunction correction is:

$$\boxed{|n^{(1)}\rangle = \sum_{m \neq n} \frac{\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle}{E_n^{(0)} - E_m^{(0)}} |m^{(0)}\rangle} \tag{17.15}$$

Physical Interpretation

This result is rich with physical meaning:

1. The perturbation mixes states. The perturbed state $|n\rangle \approx |n^{(0)}\rangle + \lambda|n^{(1)}\rangle$ is no longer a pure unperturbed eigenstate. It acquires components from other unperturbed states, with the amount of admixture given by the matrix element $\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle$ divided by the energy difference.

2. Nearby states mix more. States with energies close to $E_n^{(0)}$ (small denominator) contribute more to the correction. States far away in energy contribute less. This makes intuitive sense: a perturbation can more easily "push" a state into mixing with an energetically nearby state than a distant one.

3. The matrix element is the coupling strength. If $\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle = 0$ — that is, if the perturbation does not connect states $|n^{(0)}\rangle$ and $|m^{(0)}\rangle$ — then state $|m^{(0)}\rangle$ does not mix into state $|n\rangle$ at first order, regardless of how close the energies are. Symmetry often forces matrix elements to vanish, drastically simplifying calculations.

4. Degeneracy is catastrophic. If $E_m^{(0)} = E_n^{(0)}$ for some $m \neq n$, the denominator in (17.14) vanishes and the coefficient diverges. This is why the non-degeneracy assumption is essential. In the degenerate case (Chapter 18), we must first diagonalize $\hat{H}'$ within the degenerate subspace.

🔴 Warning — The sum in (17.15) runs over all unperturbed states $|m^{(0)}\rangle$ with $m \neq n$, including continuum states if the spectrum is not purely discrete. For bound-state problems with a continuum (like hydrogen), the sum becomes a sum over discrete states plus an integral over the continuum. In practice, the continuum contribution is often small for low-lying states.

Selection Rules and Simplification

In many problems, symmetry dramatically reduces the number of terms in the sum (17.15). A state $|m^{(0)}\rangle$ contributes to the correction only if:

$$\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle \neq 0$$

For example, if $\hat{H}'$ is an odd-parity operator (like $\hat{x}$ or $\hat{x}^3$), then it connects states of opposite parity. If $\hat{H}'$ is even-parity (like $\hat{x}^2$ or $\hat{x}^4$), it connects states of the same parity. These selection rules can often reduce an infinite sum to a handful of terms.

🔗 Connection — The ladder operators from Chapter 4/8 are enormously useful here. For the harmonic oscillator, $\hat{x} = \sqrt{\hbar/(2m\omega)}\,(\hat{a} + \hat{a}^\dagger)$ connects $|n\rangle$ only to $|n\pm 1\rangle$, while $\hat{x}^2$ connects $|n\rangle$ to $|n\rangle$, $|n\pm 2\rangle$, and so on. This turns infinite sums into finite sums.

Worked Example: First-Order State Correction for the Linear Perturbation

Returning to our QHO with $\hat{H}' = F\hat{x}$, we compute $|n^{(1)}\rangle$. The matrix elements are:

$$\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle = F\sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt{n}\,\delta_{m,n-1} + \sqrt{n+1}\,\delta_{m,n+1}\right)$$

Only two terms survive in the sum. The first-order correction to the ground state is:

$$|0^{(1)}\rangle = \frac{\langle 1^{(0)}|F\hat{x}|0^{(0)}\rangle}{E_0^{(0)} - E_1^{(0)}}|1^{(0)}\rangle = \frac{F\sqrt{\hbar/(2m\omega)}}{-\hbar\omega}|1^{(0)}\rangle = -\frac{F}{\hbar\omega}\sqrt{\frac{\hbar}{2m\omega}}|1^{(0)}\rangle$$

Since $|1^{(0)}\rangle$ has a wavefunction proportional to $x\,e^{-m\omega x^2/(2\hbar)}$, the correction $|0^{(1)}\rangle$ introduces an asymmetry: the probability distribution shifts in the direction of the applied force. This is exactly what we expect physically — the particle's equilibrium position moves to $x_0 = -F/(m\omega^2)$.

The corrected ground state (to first order in $F$) is:

$$|0\rangle \approx |0^{(0)}\rangle - \frac{F}{\hbar\omega}\sqrt{\frac{\hbar}{2m\omega}}|1^{(0)}\rangle$$

In position representation, this is a Gaussian centered slightly off the origin — the beginning of a shift toward the new equilibrium. Higher-order corrections complete the shift.

The Validity Criterion

The expansion (17.15) is valid only if the corrections are small: $|\lambda c_m^{(1)}| \ll 1$ for all $m$. This requires:

$$|\lambda\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle| \ll |E_n^{(0)} - E_m^{(0)}| \tag{17.16}$$

In words: the perturbation matrix elements must be small compared to the energy level spacings. This is the fundamental validity criterion for non-degenerate perturbation theory. When this condition is violated for even a single state $m$, the expansion for state $n$ breaks down — either because the correction is too large (strong coupling) or because the denominator vanishes (degeneracy).

17.5 Second-Order Energy Correction

The Derivation

The first-order energy correction captures the effect of the perturbation to leading order. But many important physical effects — including the Stark effect for the hydrogen ground state — vanish at first order and only appear at second order. The second-order correction turns out to involve the first-order wavefunction, which we have just calculated.

We return to the second-order equation (17.9) and act on the left with $\langle n^{(0)}|$:

$$\langle n^{(0)}|\hat{H}_0|n^{(2)}\rangle + \langle n^{(0)}|\hat{H}'|n^{(1)}\rangle = E_n^{(0)}\langle n^{(0)}|n^{(2)}\rangle + E_n^{(1)}\langle n^{(0)}|n^{(1)}\rangle + E_n^{(2)}\langle n^{(0)}|n^{(0)}\rangle$$

Using the same three facts as before: $\langle n^{(0)}|\hat{H}_0 = E_n^{(0)}\langle n^{(0)}|$, $\langle n^{(0)}|n^{(2)}\rangle = 0$ (normalization convention), and $\langle n^{(0)}|n^{(1)}\rangle = 0$. The left-hand side simplifies:

$$E_n^{(0)}\langle n^{(0)}|n^{(2)}\rangle + \langle n^{(0)}|\hat{H}'|n^{(1)}\rangle = E_n^{(0)} \cdot 0 + \langle n^{(0)}|\hat{H}'|n^{(1)}\rangle$$

The right-hand side simplifies:

$$E_n^{(0)} \cdot 0 + E_n^{(1)} \cdot 0 + E_n^{(2)} \cdot 1 = E_n^{(2)}$$

Therefore:

$$E_n^{(2)} = \langle n^{(0)}|\hat{H}'|n^{(1)}\rangle \tag{17.16}$$

Now substitute the explicit expression (17.15) for $|n^{(1)}\rangle$:

$$E_n^{(2)} = \sum_{m \neq n} \frac{\langle n^{(0)}|\hat{H}'|m^{(0)}\rangle\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle}{E_n^{(0)} - E_m^{(0)}}$$

Since $\langle n^{(0)}|\hat{H}'|m^{(0)}\rangle = \langle m^{(0)}|\hat{H}'|n^{(0)}\rangle^*$ (because $\hat{H}'$ is Hermitian), we can write:

$$\boxed{E_n^{(2)} = \sum_{m \neq n} \frac{|\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle|^2}{E_n^{(0)} - E_m^{(0)}}} \tag{17.17}$$

Properties of the Second-Order Correction

This result has several crucial properties:

The ground state is always lowered. For the ground state ($n = 0$), every term in the sum has $E_0^{(0)} - E_m^{(0)} < 0$ (since $E_0^{(0)}$ is the lowest energy), and the numerator $|\cdots|^2$ is non-negative. Therefore $E_0^{(2)} \le 0$: the second-order correction always lowers the ground-state energy. This makes physical sense — the perturbation gives the system more "room to maneuver," and the ground state takes advantage of this freedom to minimize energy.

💡 Key Insight — The fact that $E_0^{(2)} \le 0$ is actually a special case of a general principle. It follows from the variational theorem (Chapter 19): the exact ground-state energy must be lower than or equal to the expectation value $\langle 0^{(0)}|\hat{H}|0^{(0)}\rangle = E_0^{(0)} + \lambda E_0^{(1)}$, so the second-order (and higher) corrections must bring the energy down.

Excited states can go either way. For excited states, some terms have positive denominators (from states below) and some have negative denominators (from states above). The sign of $E_n^{(2)}$ depends on which contributions dominate — it can be positive, negative, or zero.

Level repulsion. Consider two states $|n^{(0)}\rangle$ and $|m^{(0)}\rangle$ with energies $E_n^{(0)} < E_m^{(0)}$. The second-order correction to $E_n$ from the coupling to $|m^{(0)}\rangle$ is negative (pushes $n$ down), while the correction to $E_m$ from coupling to $|n^{(0)}\rangle$ is positive (pushes $m$ up). The two levels repel each other. This is a general phenomenon in quantum mechanics and is particularly important in molecular orbital theory and band structure calculations.

⚠️ Common Misconception — Students sometimes assume that only the nearest unperturbed state matters. In fact, all states contribute, and sometimes the cumulative effect of many distant states exceeds the contribution of a single nearby state. Always consider the full sum before truncating.

The Complete Energy to Second Order

Collecting our results, the energy of state $|n\rangle$ through second order is:

$$E_n = E_n^{(0)} + \lambda\langle n^{(0)}|\hat{H}'|n^{(0)}\rangle + \lambda^2\sum_{m \neq n} \frac{|\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle|^2}{E_n^{(0)} - E_m^{(0)}} + O(\lambda^3) \tag{17.18}$$

When $\lambda = 1$ (the physical case), this becomes:

$$E_n \approx E_n^{(0)} + \langle n^{(0)}|\hat{H}'|n^{(0)}\rangle + \sum_{m \neq n} \frac{|\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle|^2}{E_n^{(0)} - E_m^{(0)}} \tag{17.19}$$

Worked Example: Two-State System

To build concrete intuition, consider a two-state system with unperturbed energies $E_1^{(0)} = 0$ and $E_2^{(0)} = \Delta$ (with $\Delta > 0$), and a perturbation that connects them: $\langle 1^{(0)}|\hat{H}'|2^{(0)}\rangle = V$ (real). The diagonal elements are zero: $\langle 1^{(0)}|\hat{H}'|1^{(0)}\rangle = \langle 2^{(0)}|\hat{H}'|2^{(0)}\rangle = 0$.

Perturbation theory results:

• First-order energies: $E_1^{(1)} = 0$, $E_2^{(1)} = 0$.
• Second-order energies: $E_1^{(2)} = |V|^2/(0 - \Delta) = -V^2/\Delta$ and $E_2^{(2)} = |V|^2/(\Delta - 0) = +V^2/\Delta$.

The lower level is pushed down, the upper level is pushed up: level repulsion. The corrected energies are:

$$E_1 \approx -\frac{V^2}{\Delta}, \qquad E_2 \approx \Delta + \frac{V^2}{\Delta}$$

Exact solution: The $2\times 2$ Hamiltonian matrix is:

$$H = \begin{pmatrix} 0 & V \\ V & \Delta \end{pmatrix}$$

The exact eigenvalues are:

$$E_{\pm} = \frac{\Delta}{2} \pm \frac{1}{2}\sqrt{\Delta^2 + 4V^2}$$

Expanding to second order in $V/\Delta$:

$$E_- = \frac{\Delta}{2} - \frac{\Delta}{2}\sqrt{1 + 4V^2/\Delta^2} \approx \frac{\Delta}{2} - \frac{\Delta}{2}\left(1 + \frac{2V^2}{\Delta^2}\right) = -\frac{V^2}{\Delta}$$

$$E_+ = \frac{\Delta}{2} + \frac{\Delta}{2}\sqrt{1 + 4V^2/\Delta^2} \approx \Delta + \frac{V^2}{\Delta}$$

These agree exactly with the perturbation theory results. ✓

This two-state example also reveals when perturbation theory breaks down: when $V/\Delta \gtrsim 1$, the exact eigenvalues deviate significantly from the perturbative approximation. The exact eigenvalues never cross (this is the avoided crossing or anticrossing phenomenon), while perturbation theory cannot capture this nonlinear behavior at any finite order.

📊 By the Numbers — Avoided crossings are ubiquitous in quantum mechanics. In molecular physics, the Born-Oppenheimer potential energy curves for diatomic molecules exhibit avoided crossings whenever two electronic states are coupled by non-adiabatic interactions. These crossings govern photodissociation, predissociation, and many other photochemical processes. Understanding them begins with the two-state perturbation model we have just analyzed.

✅ Checkpoint — Make sure you can: - Derive (17.17) from (17.16) and (17.15) - Explain why the second-order correction always lowers the ground-state energy - State what "level repulsion" means and give a physical example - Identify what goes wrong when two unperturbed energies are equal

17.6 Application: The Anharmonic Oscillator

Setting Up the Problem

The quantum harmonic oscillator has perfectly equally spaced energy levels, a feature that makes it exactly solvable but physically unrealistic for large excitations. Real molecular potentials are anharmonic — the restoring force is not exactly proportional to displacement. The simplest anharmonic correction is a quartic term:

$$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2 + \lambda\hat{x}^4 \tag{17.20}$$

Here $\hat{H}_0 = \hat{p}^2/(2m) + m\omega^2\hat{x}^2/2$ is the harmonic oscillator, and $\hat{H}' = \hat{x}^4$.

🔗 Connection — We could also consider a cubic perturbation $\hat{H}' = \hat{x}^3$, which is physically relevant for real molecular potentials (the Morse potential has a cubic leading correction). However, $\hat{x}^3$ has odd parity, so its first-order correction vanishes for all QHO states (which have definite parity). The cubic term contributes only at second order. The quartic term $\hat{x}^4$ has even parity and gives a nonzero first-order correction, making it the more instructive example.

First-Order Energy Correction

We need $E_n^{(1)} = \langle n^{(0)}|\hat{x}^4|n^{(0)}\rangle$. The key is to express $\hat{x}$ in terms of ladder operators:

$$\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}(\hat{a} + \hat{a}^\dagger) \tag{17.21}$$

Therefore:

$$\hat{x}^4 = \left(\frac{\hbar}{2m\omega}\right)^2 (\hat{a} + \hat{a}^\dagger)^4 \tag{17.22}$$

We need to expand $(\hat{a} + \hat{a}^\dagger)^4$ and compute its expectation value in $|n\rangle$. This requires patient application of the commutation relation $[\hat{a}, \hat{a}^\dagger] = 1$ and the action of ladder operators:

$$\hat{a}|n\rangle = \sqrt{n}|n-1\rangle, \qquad \hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$$

The key insight is that $\langle n|(\hat{a}+\hat{a}^\dagger)^4|n\rangle$ only receives contributions from terms that contain equal numbers of $\hat{a}$'s and $\hat{a}^\dagger$'s — otherwise the result would be orthogonal to $|n\rangle$. After careful expansion (or using the technique of normal ordering), one finds:

$$\langle n|(\hat{a}+\hat{a}^\dagger)^4|n\rangle = 6n^2 + 6n + 3 \tag{17.23}$$

Let us verify this for the ground state ($n = 0$). We need $\langle 0|(\hat{a}+\hat{a}^\dagger)^4|0\rangle$. Expanding:

$$(\hat{a}+\hat{a}^\dagger)^4 = \hat{a}^4 + \hat{a}^3\hat{a}^\dagger + \hat{a}^2\hat{a}^{\dagger 2} + \hat{a}^2\hat{a}^\dagger\hat{a} + \hat{a}\hat{a}^{\dagger}\hat{a}^2 + \hat{a}\hat{a}^\dagger\hat{a}\hat{a}^\dagger + \hat{a}\hat{a}^{\dagger 2}\hat{a} + \cdots$$

This gets complicated fast. A more efficient approach is to compute stepwise:

$$(\hat{a}+\hat{a}^\dagger)^2|0\rangle = (\hat{a}^2 + \hat{a}\hat{a}^\dagger + \hat{a}^\dagger\hat{a} + \hat{a}^{\dagger 2})|0\rangle$$

Since $\hat{a}|0\rangle = 0$: $\hat{a}^2|0\rangle = 0$, $\hat{a}^\dagger\hat{a}|0\rangle = 0$, $\hat{a}\hat{a}^\dagger|0\rangle = (\hat{a}^\dagger\hat{a} + 1)|0\rangle = |0\rangle$, and $\hat{a}^{\dagger 2}|0\rangle = \sqrt{2}|2\rangle$.

So $(\hat{a}+\hat{a}^\dagger)^2|0\rangle = |0\rangle + \sqrt{2}|2\rangle$.

Now apply $(\hat{a}+\hat{a}^\dagger)^2$ again:

$$(\hat{a}+\hat{a}^\dagger)^2(|0\rangle + \sqrt{2}|2\rangle) = (\hat{a}+\hat{a}^\dagger)^2|0\rangle + \sqrt{2}(\hat{a}+\hat{a}^\dagger)^2|2\rangle$$

We already know the first piece: $|0\rangle + \sqrt{2}|2\rangle$. For the second:

$$(\hat{a}+\hat{a}^\dagger)^2|2\rangle = (\hat{a}^2 + \hat{a}\hat{a}^\dagger + \hat{a}^\dagger\hat{a} + \hat{a}^{\dagger 2})|2\rangle$$

Computing each term: $\hat{a}^2|2\rangle = \sqrt{2}|0\rangle$, $\hat{a}\hat{a}^\dagger|2\rangle = \hat{a}\sqrt{3}|3\rangle = \sqrt{6}|2\rangle \to$ wait, let us be more careful. $\hat{a}\hat{a}^\dagger|2\rangle = (\hat{a}^\dagger\hat{a} + 1)|2\rangle = (2+1)|2\rangle = 3|2\rangle$. And $\hat{a}^\dagger\hat{a}|2\rangle = 2|2\rangle$. And $\hat{a}^{\dagger 2}|2\rangle = \hat{a}^\dagger\sqrt{3}|3\rangle = \sqrt{3}\sqrt{4}|4\rangle = 2\sqrt{3}|4\rangle$.

So $(\hat{a}+\hat{a}^\dagger)^2|2\rangle = \sqrt{2}|0\rangle + 5|2\rangle + 2\sqrt{3}|4\rangle$.

Putting it all together:

$$(\hat{a}+\hat{a}^\dagger)^4|0\rangle = (|0\rangle + \sqrt{2}|2\rangle) + \sqrt{2}(\sqrt{2}|0\rangle + 5|2\rangle + 2\sqrt{3}|4\rangle) = 3|0\rangle + 6\sqrt{2}|2\rangle + 2\sqrt{6}|4\rangle$$

Therefore: $\langle 0|(\hat{a}+\hat{a}^\dagger)^4|0\rangle = 3$, which agrees with $6(0)^2 + 6(0) + 3 = 3$. ✓

The Result

The first-order energy correction for the anharmonic oscillator is:

$$\boxed{E_n^{(1)} = \lambda\left(\frac{\hbar}{2m\omega}\right)^2(6n^2 + 6n + 3)} \tag{17.24}$$

For the ground state:

$$E_0^{(1)} = 3\lambda\left(\frac{\hbar}{2m\omega}\right)^2 = \frac{3\lambda\hbar^2}{4m^2\omega^2} \tag{17.25}$$

Several features are worth noting:

1. The correction is always positive for $\lambda > 0$ (the quartic term raises the potential everywhere), consistent with physical intuition.

2. The correction grows with $n$. Higher-lying states sample larger regions of $x$-space where the $x^4$ term is more important, so the quartic correction is larger. This means the energy levels are no longer equally spaced — the spacing increases with $n$.

3. The equal spacing is broken quadratically. The $6n^2$ term dominates for large $n$, so the anharmonic correction grows as $n^2$, compared to the unperturbed energy which grows linearly in $n$.

📊 By the Numbers — For a CO molecule modeled with $\omega = 4.07 \times 10^{14}$ rad/s and a quartic anharmonicity parameter $\lambda \approx 1.2 \times 10^{43}$ J/m⁴, the first-order correction to the ground state is approximately $0.3\%$ of $\hbar\omega$. This is small enough for perturbation theory to be valid, yet large enough to be spectroscopically measurable — precisely the regime where perturbation theory shines.

Second-Order Correction for the Anharmonic Oscillator

Let us also compute $E_0^{(2)}$ for the ground state. We need the matrix elements $\langle m^{(0)}|\hat{x}^4|0^{(0)}\rangle$ for $m \neq 0$. From the expansion we computed in the first-order verification:

$$(\hat{a}+\hat{a}^\dagger)^4|0\rangle = 3|0\rangle + 6\sqrt{2}|2\rangle + 2\sqrt{6}|4\rangle$$

Therefore, multiplying by $(\hbar/(2m\omega))^2$:

$$\langle 2|\hat{x}^4|0\rangle = \left(\frac{\hbar}{2m\omega}\right)^2 \cdot 6\sqrt{2}, \qquad \langle 4|\hat{x}^4|0\rangle = \left(\frac{\hbar}{2m\omega}\right)^2 \cdot 2\sqrt{6}$$

In natural units ($\hbar = m = \omega = 1$), $(\hbar/(2m\omega))^2 = 1/4$, so $\langle 2|\hat{x}^4|0\rangle = 3\sqrt{2}/2$ and $\langle 4|\hat{x}^4|0\rangle = \sqrt{6}/2$.

The second-order correction is:

$$E_0^{(2)} = \lambda^2\left[\frac{|\langle 2|\hat{x}^4|0\rangle|^2}{E_0^{(0)} - E_2^{(0)}} + \frac{|\langle 4|\hat{x}^4|0\rangle|^2}{E_0^{(0)} - E_4^{(0)}}\right]$$

$$= \lambda^2\left[\frac{(3\sqrt{2}/2)^2}{1/2 - 5/2} + \frac{(\sqrt{6}/2)^2}{1/2 - 9/2}\right] = \lambda^2\left[\frac{9/2}{-2} + \frac{3/2}{-4}\right] = \lambda^2\left[-\frac{9}{4} - \frac{3}{8}\right] = -\frac{21}{8}\lambda^2$$

So the ground-state energy through second order is (in natural units):

$$E_0 \approx \frac{1}{2} + \frac{3}{4}\lambda - \frac{21}{8}\lambda^2$$

For $\lambda = 0.1$: $E_0 \approx 0.5 + 0.075 - 0.02625 = 0.54875$. Numerical diagonalization gives $E_0 = 0.5592$, so the second-order result is accurate to about 2% — much better than the 3.5% error of first-order alone. For $\lambda = 0.01$: $E_0 \approx 0.5 + 0.0075 - 0.0002625 = 0.5072375$, while the exact value is $E_0 = 0.50744$, giving an error of only $0.04\%$.

Comparison: Perturbation Theory vs. Exact Diagonalization

The table below summarizes the accuracy of perturbation theory for the anharmonic oscillator ground state at different coupling strengths (natural units):

The message is clear: perturbation theory works beautifully for $\lambda \lesssim 0.1$ and catastrophically fails for $\lambda \sim 1$.

17.7 Application: The Stark Effect for the Hydrogen Ground State

Physical Setup

When a hydrogen atom is placed in a uniform external electric field $\mathbf{\mathcal{E}} = \mathcal{E}\hat{z}$, the electron experiences an additional potential energy:

$$\hat{H}' = e\mathcal{E}z = e\mathcal{E}r\cos\theta \tag{17.26}$$

(Here $e > 0$ is the magnitude of the electron charge, and we use the convention that the electron's charge is $-e$, so the perturbation is $+e\mathcal{E}z$ representing the potential energy of the electron in the field.)

This is the Stark effect, named after Johannes Stark, who observed the splitting of hydrogen spectral lines in electric fields in 1913 — the same year Bohr proposed his model.

🔵 Historical Note — Stark's observation of spectral line splitting in electric fields earned him the 1919 Nobel Prize in Physics. The Stark effect was one of the early triumphs of quantum mechanics, as Schrödinger showed in 1926 that his equation could quantitatively explain the observed splittings. Stark himself later became an enthusiastic Nazi and attempted to purge "Jewish physics" from German universities — a cautionary tale about the relationship between scientific achievement and moral character.

First-Order Correction: Why It Vanishes

For the hydrogen ground state $|1,0,0\rangle$ (with $n=1, l=0, m=0$), the first-order energy correction is:

$$E_1^{(1)} = \langle 1,0,0|\,e\mathcal{E}r\cos\theta\,|1,0,0\rangle = e\mathcal{E}\langle 1,0,0|\,r\cos\theta\,|1,0,0\rangle \tag{17.27}$$

The ground state wavefunction is:

$$\psi_{100}(r,\theta,\phi) = \frac{1}{\sqrt{\pi}}\left(\frac{1}{a_0}\right)^{3/2} e^{-r/a_0}$$

This function is spherically symmetric — it has no angular dependence. The perturbation $r\cos\theta$ has the angular dependence of $Y_1^0(\theta,\phi) \propto \cos\theta$, which is odd under the parity transformation $\mathbf{r} \to -\mathbf{r}$ (i.e., $\theta \to \pi - \theta$). Since $|\psi_{100}|^2$ is even under parity, the integral:

$$\langle 1,0,0|\,r\cos\theta\,|1,0,0\rangle = \int_0^\infty r^2 dr \int_0^\pi \sin\theta\,d\theta \int_0^{2\pi} d\phi\, |\psi_{100}|^2 \cdot r\cos\theta$$

vanishes by symmetry. The $\cos\theta$ factor makes the integrand odd in $\theta$, and the integral over $\theta$ from $0$ to $\pi$ gives zero.

$$E_1^{(1)} = 0 \tag{17.28}$$

💡 Key Insight — The vanishing of the first-order Stark effect for the hydrogen ground state is a consequence of parity symmetry. The ground state has even parity ($l = 0$), the perturbation $r\cos\theta \propto z$ has odd parity, so the matrix element vanishes. This is not special to hydrogen — any non-degenerate state with definite parity has zero first-order Stark effect. The physical statement is profound: a spherically symmetric charge distribution has no permanent electric dipole moment.

⚠️ Common Misconception — The vanishing of the first-order correction does NOT mean the electric field has no effect. It means we need to go to second order. The field induces a dipole moment in the atom — it distorts the otherwise spherical charge distribution — and this induced dipole then interacts with the field. This is a second-order effect.

Second-Order Correction: Atomic Polarizability

Since the first-order correction vanishes, the leading effect of the electric field on the ground-state energy is the second-order correction:

$$E_1^{(2)} = \sum_{n,l,m \neq 1,0,0} \frac{|\langle n,l,m|\,e\mathcal{E}r\cos\theta\,|1,0,0\rangle|^2}{E_1^{(0)} - E_{n}^{(0)}} \tag{17.29}$$

where the sum runs over all states of hydrogen except the ground state (including the continuum).

Selection rules dramatically simplify this. The matrix element $\langle n,l,m|r\cos\theta|1,0,0\rangle$ is nonzero only if $\Delta l = \pm 1$ and $\Delta m = 0$. Since the ground state has $l = 0$, $m = 0$, the only contributing states have $l = 1$, $m = 0$: the $|n,1,0\rangle$ states.

$$E_1^{(2)} = e^2\mathcal{E}^2 \sum_{n=2}^{\infty} \frac{|\langle n,1,0|\,r\cos\theta\,|1,0,0\rangle|^2}{E_1^{(0)} - E_n^{(0)}} + \text{continuum} \tag{17.30}$$

Bounding the Result: Dalgarno-Lewis and the Exact Answer

Computing the full sum (including the continuum) is technically demanding. A clever alternative is the Dalgarno-Lewis method, which replaces the infinite sum with the solution of an inhomogeneous differential equation. Without going through the full calculation (see Further Reading), the exact result is:

$$E_1^{(2)} = -\frac{9}{4}a_0^3 e^2\mathcal{E}^2 \cdot \frac{1}{4\pi\epsilon_0} = -\frac{9}{4}\frac{a_0^3\mathcal{E}^2}{4\pi\epsilon_0/e^2}$$

It is conventional to write this as:

$$E_1^{(2)} = -\frac{1}{2}\alpha\mathcal{E}^2 \tag{17.31}$$

where $\alpha$ is the static polarizability of the hydrogen ground state:

$$\boxed{\alpha = \frac{9}{2}(4\pi\epsilon_0)a_0^3 = \frac{9a_0^3}{2} \text{ (in Gaussian units)}} \tag{17.32}$$

The energy shift is proportional to $\mathcal{E}^2$, not $\mathcal{E}$ — this is the hallmark of an induced (rather than permanent) dipole moment. The coefficient $\alpha$ tells us how "squashy" the atom is: how easily the electric field can distort the electronic charge distribution.

📊 By the Numbers — The polarizability of hydrogen is $\alpha/(4\pi\epsilon_0) = 4.5 a_0^3 = 0.667\,\text{\AA}^3$. For an electric field of $\mathcal{E} = 10^7$ V/m (achievable in the laboratory), the energy shift is $\Delta E \approx 2 \times 10^{-8}$ eV — small but measurable spectroscopically. Compare this to the ground-state energy of $13.6$ eV: the perturbation is a part in $10^9$, well within the perturbative regime.

A Useful Bound

Even without computing the exact sum, we can get a useful upper bound on $|E_1^{(2)}|$. The key trick is to note that all excited states have $E_n^{(0)} \ge E_2^{(0)}$, so:

$$|E_n^{(0)} - E_1^{(0)}| \ge E_2^{(0)} - E_1^{(0)} = \frac{3}{4} \cdot 13.6\,\text{eV} = 10.2\,\text{eV}$$

Therefore:

$$|E_1^{(2)}| \le \frac{e^2\mathcal{E}^2}{E_2^{(0)} - E_1^{(0)}} \sum_{n \neq 1} |\langle n|r\cos\theta|1,0,0\rangle|^2$$

The sum can be evaluated using the closure relation (completeness):

$$\sum_{n \neq 1} |\langle n|r\cos\theta|1,0,0\rangle|^2 = \langle 1,0,0|(r\cos\theta)^2|1,0,0\rangle - |\langle 1,0,0|r\cos\theta|1,0,0\rangle|^2$$

The second term vanishes (as we showed above). The first term is:

$$\langle 1,0,0|r^2\cos^2\theta|1,0,0\rangle = \langle r^2\rangle_{100} \cdot \langle\cos^2\theta\rangle = 3a_0^2 \cdot \frac{1}{3} = a_0^2$$

This gives the bound $|E_1^{(2)}| \le e^2\mathcal{E}^2 a_0^2 / (10.2\,\text{eV})$, corresponding to a polarizability bound of $\alpha/(4\pi\epsilon_0) \le 2a_0^3/10.2 \cdot 13.6 = 2.67 a_0^3$. The exact result $4.5 a_0^3$ is larger because the bound replaces all energy denominators with the largest one, which overestimates them.

The Dalgarno-Lewis Method: A Powerful Shortcut

Computing the infinite sum in (17.29) directly — including the continuum — is formidable. Dalgarno and Lewis (1955) introduced an elegant alternative that avoids the sum entirely.

The idea is to find a function $|F\rangle$ that satisfies:

$$(\hat{H}_0 - E_1^{(0)})|F\rangle = -(r\cos\theta - \langle r\cos\theta\rangle)|1,0,0\rangle = -r\cos\theta\,|1,0,0\rangle$$

where the last equality uses $\langle r\cos\theta\rangle = 0$. If we can find $|F\rangle$, then the second-order energy is simply:

$$E_1^{(2)} = e^2\mathcal{E}^2\langle 1,0,0|r\cos\theta|F\rangle$$

The key insight is that in position representation, $|F\rangle = f(r)\cos\theta\, \psi_{100}(r)$ for some radial function $f(r)$. Substituting into the differential equation and solving (which reduces to a first-order ODE for $f(r)$), one finds:

$$f(r) = -a_0\left(\frac{r}{a_0} + \frac{r^2}{2a_0^2}\right)$$

The integral $\langle 1,0,0|r\cos\theta|F\rangle$ is then a standard radial integral, yielding the exact result $\alpha/(4\pi\epsilon_0) = (9/2)a_0^3$. This method bypasses the infinite sum entirely — a beautiful example of mathematical efficiency.

🧪 Experiment — The Stark effect has been measured with extraordinary precision using atomic beam techniques and laser spectroscopy. For hydrogen, the measured polarizability agrees with the theoretical value to better than $0.1\%$. This is a beautiful confirmation of second-order perturbation theory. Modern measurements using Rydberg atoms in electric fields have reached precisions of parts per million, providing stringent tests of quantum electrodynamic corrections to the polarizability.

The Induced Dipole Moment

The energy shift $\Delta E = -\frac{1}{2}\alpha\mathcal{E}^2$ has a direct physical interpretation in terms of the induced dipole moment. The electric field distorts the spherically symmetric ground-state charge distribution, creating a dipole moment:

$$\langle d_z\rangle = -\frac{\partial(\Delta E)}{\partial\mathcal{E}} = \alpha\mathcal{E}$$

This is the quantum-mechanical version of classical polarization: the applied field induces a dipole proportional to the field strength, with the polarizability $\alpha$ as the proportionality constant. The factor of $1/2$ in the energy formula arises because work must be done to create the dipole — it is the same factor that appears in the energy of a capacitor.

For hydrogen in a field of $\mathcal{E} = 10^7$ V/m, the induced dipole moment is $\langle d_z\rangle \approx 5 \times 10^{-40}$ C$\cdot$m $\approx 1.5 \times 10^{-10} ea_0$ — an extraordinarily tiny displacement. The electron cloud shifts by a fraction of a femtometer. Yet this minuscule effect is readily measured spectroscopically.

✅ Checkpoint — Make sure you can: - Explain why the first-order Stark effect vanishes for the hydrogen ground state (two arguments: parity, and the physical statement about permanent dipole moments) - Write the general formula for the second-order correction and identify the selection rules that simplify the sum - Explain why the energy shift is proportional to $\mathcal{E}^2$ and what this tells us about induced vs. permanent dipoles

17.8 Convergence: When Perturbation Theory Works — and When It Fails

The Convergence Question

We have been treating the perturbation expansion as if it converges — as if adding more terms brings us closer to the true answer. But does it? The honest answer is: often it does not. The perturbation series is frequently an asymptotic series rather than a convergent one, meaning that the first few terms give good approximations but the series eventually diverges if pushed to high order.

This is not a mathematical curiosity — it is a fundamental feature of quantum mechanics with deep physical roots.

The Radius of Convergence Problem

Consider the anharmonic oscillator $\hat{H} = \hat{H}_0 + \lambda\hat{x}^4$. For $\lambda > 0$, the potential rises steeply at large $|x|$ and the Hamiltonian has a discrete spectrum — the system is well-behaved. But for $\lambda < 0$, the potential $V(x) = \frac{1}{2}m\omega^2 x^2 + \lambda x^4$ is unbounded below: the particle can tunnel to arbitrarily negative energies through the potential barrier. This means the energy levels do not exist as stable states for any $\lambda < 0$, no matter how small $|\lambda|$ is.

This has a devastating consequence for the perturbation series. A power series $\sum_k a_k \lambda^k$ converges within a circle of radius $R$ in the complex $\lambda$-plane, and $R$ is determined by the nearest singularity to the origin. Since the physics is singular at $\lambda = 0^-$ (the energy levels cease to exist), the radius of convergence is $R = 0$.

🔴 Warning — The perturbation series for the anharmonic oscillator has zero radius of convergence. It diverges for any nonzero $\lambda$, no matter how small. Yet the first few terms give excellent numerical results for small $\lambda$. This is the hallmark of an asymptotic series: useful in practice despite being formally divergent.

Asymptotic Series: Useful Despite Divergence

An asymptotic series $\sum_{k=0}^{N} a_k \lambda^k$ for a function $f(\lambda)$ satisfies:

$$\left|f(\lambda) - \sum_{k=0}^{N} a_k \lambda^k\right| < C_{N+1} |\lambda|^{N+1}$$

for sufficiently small $\lambda$. The error after $N$ terms is bounded by the next term — but only for fixed $N$ as $\lambda \to 0$. If you fix $\lambda$ and let $N \to \infty$, the series eventually diverges.

The practical consequence: there is an optimal truncation order $N^*(\lambda)$ beyond which adding more terms makes the approximation worse, not better. For the anharmonic oscillator, $N^* \sim 1/\lambda$ (roughly), so for $\lambda = 0.01$, the first $\sim 100$ terms improve the answer, but beyond that the series deteriorates.

When Does Perturbation Theory Work?

Perturbation theory gives reliable results when:

1. The perturbation is genuinely small. Specifically, $|\lambda\langle m^{(0)}|\hat{H}'|n^{(0)}\rangle| \ll |E_n^{(0)} - E_m^{(0)}|$ for all relevant states $m$. The perturbation matrix elements must be small compared to energy level spacings.

2. The perturbation does not qualitatively change the physics. If the perturbation introduces new bound states, opens a continuum channel, or changes the topology of the potential, perturbation theory will fail at any order.

3. There is no degeneracy or near-degeneracy. If two unperturbed levels are separated by an energy gap much smaller than the perturbation matrix element connecting them, non-degenerate perturbation theory fails. (Degenerate perturbation theory, Chapter 18, handles exact degeneracy; for near-degeneracy, one must treat the nearly degenerate subspace exactly.)

Example: When It Fails Spectacularly

Consider a double-well potential: $V(x) = \lambda(x^2 - a^2)^2$. For large $\lambda$, each well is approximately a harmonic oscillator, and the low-lying states are localized in one well or the other. The energy splitting between the symmetric and antisymmetric combinations is exponentially small: $\Delta E \sim e^{-S/\hbar}$, where $S$ is the instanton action (related to tunneling through the barrier).

No finite order of perturbation theory in $\lambda$ can reproduce an exponentially small splitting. The splitting is a non-perturbative effect — it is invisible to the perturbation expansion around either well. This is one of the most important lessons in theoretical physics: some physical effects lie entirely outside the reach of perturbation theory.

💡 Key Insight — The existence of non-perturbative effects does not invalidate perturbation theory. It means that perturbation theory captures only part of the physics — the part that varies as a power of the coupling constant. Effects that vary as $e^{-c/\lambda}$ (exponentially small in $1/\lambda$) are missed entirely. Understanding which effects are perturbative and which are not is one of the central skills of theoretical physics.

Improving Convergence: A Brief Survey

Several techniques exist to extract more information from a divergent perturbation series:

1. Padé approximants: Replace the polynomial $\sum_k a_k \lambda^k$ with a ratio of polynomials $P_N(\lambda)/Q_M(\lambda)$ that matches the same Taylor coefficients. Padé approximants often converge even when the Taylor series does not.

2. Borel summation: Transform the series $\sum a_k \lambda^k$ into $\sum (a_k/k!) t^k$, which often has better convergence. The original function is recovered by an integral transform.

3. Variational perturbation theory: Combine the perturbative expansion with a variational optimization of the unperturbed Hamiltonian. This can dramatically improve convergence.

4. Resurgence: A modern mathematical framework that systematically connects the divergence of the perturbation series to non-perturbative effects, allowing both to be computed from the same data.

⚖️ Interpretation — The fact that perturbation series in quantum mechanics are typically divergent has been interpreted in different ways. Some see it as evidence that the theory is fundamentally incomplete. Others (the modern view) see it as evidence that the theory contains more information than any single expansion can capture, and that the "trans-series" combining perturbative and non-perturbative contributions gives the complete answer. This is an active area of mathematical physics research.

A Practical Flowchart for Perturbation Problems

When confronting a perturbation problem, the following sequence of questions will guide you:

1. Is the spectrum of $\hat{H}_0$ non-degenerate for the state of interest? If not, use degenerate perturbation theory (Chapter 18). If nearly degenerate (energy gap comparable to perturbation matrix elements), treat the near-degenerate subspace exactly.

2. Does $\hat{H}'$ have any symmetry? If $\hat{H}'$ has definite parity opposite to the state's parity, $E_n^{(1)} = 0$ and you need second order. Selection rules from angular momentum conservation, time-reversal symmetry, or other symmetries may eliminate large classes of matrix elements.

3. Is $E_n^{(1)}$ sufficient, or do you need $E_n^{(2)}$? If $E_n^{(1)} \neq 0$ and the perturbation is small, first order often suffices. If $E_n^{(1)} = 0$ (as in the Stark effect), second order is the leading term.

4. Can you evaluate the sum? For the harmonic oscillator, ladder operators reduce the sum to a few terms. For hydrogen, selection rules and the Dalgarno-Lewis method help. For more complex systems, numerical evaluation may be necessary.

5. Is the result consistent? Check that $|E_n^{(2)}| \ll |E_n^{(1)}|$ (when $E_n^{(1)} \neq 0$) or that $|E_n^{(2)}/E_n^{(0)}| \ll 1$. If not, perturbation theory may not be reliable for your problem.

17.9 Summary and Progressive Project

What We Have Accomplished

In this chapter, we developed the complete framework of non-degenerate time-independent perturbation theory. Starting from the decomposition $\hat{H} = \hat{H}_0 + \lambda\hat{H}'$, we derived:

We applied these to two canonical problems: - The anharmonic oscillator ($\hat{H}' = \hat{x}^4$): found that the quartic perturbation raises all energy levels, breaks the equal spacing, and grows as $n^2$ for large $n$. - The Stark effect for hydrogen: showed that the first-order correction vanishes by parity, and the second-order correction gives the atomic polarizability $\alpha = (9/2)(4\pi\epsilon_0)a_0^3$.

We also confronted the deep question of convergence, learning that perturbation series are typically asymptotic rather than convergent, and that non-perturbative effects lie beyond the reach of any finite-order calculation.

The Threshold Concept

💡 Key Insight — The threshold concept of this chapter is: Perturbation theory connects the solvable to the realistic. Without it, quantum mechanics would be limited to a handful of idealized models. With it, we can systematically compute the properties of atoms in fields, molecules with realistic potentials, and electrons in solids. The gap between the textbook and the laboratory is bridged not by finding new exact solutions, but by learning to perturb the ones we have.

🔗 Connection — This idea — that approximation is how real physics gets done — is Theme 4 of this textbook. It will recur in Chapter 18 (degenerate perturbation), Chapter 19 (variational method), Chapter 20 (WKB), Chapter 21 (time-dependent perturbation), Chapter 22 (scattering), and in the capstone projects.

Looking Ahead

In Chapter 18, we confront the complication we have been carefully avoiding: degeneracy. The hydrogen atom's $n = 2$ level is four-fold degenerate, and applying an electric field produces a first-order splitting that cannot be captured by the formulas in this chapter. The technique of degenerate perturbation theory — diagonalizing $\hat{H}'$ within the degenerate subspace — is one of the most powerful tools in quantum mechanics.

Progressive Project: Quantum Toolkit v0.17

Module: perturbation.py

Add a perturbation theory engine to your growing quantum simulation toolkit. The module should include:

1. perturbation_1st(H0_eigvals, H0_eigvecs, H_prime, n) — Given the eigenvalues and eigenvectors of $\hat{H}_0$ and the matrix representation of $\hat{H}'$, compute the first-order energy correction $E_n^{(1)}$ and the first-order state correction $|n^{(1)}\rangle$ for state $n$.

2. perturbation_2nd(H0_eigvals, H0_eigvecs, H_prime, n) — Compute the second-order energy correction $E_n^{(2)}$.

3. energy_correction(H0_eigvals, H0_eigvecs, H_prime, n, order=2) — Wrapper function that returns the energy corrected to the specified order.

4. anharmonic_oscillator(n_max, lambda_param, n_states=5) — Apply perturbation theory to the QHO with $\hat{H}' = \hat{x}^4$ and compare with numerical diagonalization.

See code/project-checkpoint.py for the full implementation and code/example-01-perturbation.py for worked examples with visualization.

Verification: Your anharmonic_oscillator() function should reproduce the analytical result $E_0^{(1)} = 3\lambda(\hbar/2m\omega)^2$ for the ground state (in natural units with $m = \omega = \hbar = 1$: $E_0^{(1)} = 3\lambda/4$). Compare with numerical exact diagonalization for $\lambda = 0.01, 0.1, 1.0$ and observe where perturbation theory begins to fail.

🔗 Connection — This module builds on your Operator, Ket, and eigenvalue solver from Chapters 6, 8, and 9. In Chapter 18, you will extend it to handle degenerate perturbation theory. In Chapter 19, you will compare perturbation theory results against the variational method for the same systems.

Spaced Review: Concepts from Earlier Chapters

This chapter draws heavily on material from earlier parts of the textbook. Take a moment to verify your understanding of these prerequisites:

From Chapter 4 (QHO): Can you write the energy levels $E_n = (n+1/2)\hbar\omega$, the ladder operator actions $\hat{a}|n\rangle = \sqrt{n}|n-1\rangle$ and $\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$, and express $\hat{x}$ and $\hat{p}$ in terms of ladder operators? These are essential for the anharmonic oscillator calculation.

From Chapter 5 (Hydrogen): Can you write the ground-state wavefunction $\psi_{100}$, the energy formula $E_n = -13.6\,\text{eV}/n^2$, and identify the quantum numbers $n, l, m$? Can you evaluate simple expectation values like $\langle r\rangle$ and $\langle r^2\rangle$ for the ground state? These are essential for the Stark effect.

From Chapter 8 (Dirac Notation): Can you work fluently with bra-ket notation, inner products, outer products, and the completeness relation $\sum_n |n\rangle\langle n| = \hat{1}$? The entire perturbation expansion rests on expanding unknown states in a known complete basis.

From Chapter 9 (Eigenvalue Problems): Do you understand the spectral decomposition of a Hermitian operator, and why the eigenstates form a complete orthonormal set? The non-degeneracy assumption — that all eigenvalues are distinct — is a condition on the spectral decomposition of $\hat{H}_0$.

Key Vocabulary Introduced in This Chapter